Characterizing Cycle Partition in 2-Row Bulgarian Solitaire

Transcription

1 Saint Peter s University Honors Thesis Characterizing Cycle Partition in 2-Row Bulgarian Solitaire Author: Sabin K Pradhan Advisor: Dr Brian Hopkins A thesis submitted in partial fulfillment of the requirements for a baccalaureate degree in Mathematics in Cursu Honorum Submitted to: The Honors Program, Saint Peter s University May 10, 2016

2 Abstract This paper studies cyclic partitions under the operation 2-row Bulgarian solitaire We develop tools such as block notation to make characterizing cyclic partitions easier Using these blocks, we see that cyclic partition under 2-row Bulgarian solitaire have independently cycling diagonals satisfying one of four conditions We conclude with an enumeration results that allow us to calculate the number of cyclic partitions for a given integer n 1

3 Acknowledgment I would like to take this opportunity to thank a number of people without whom this thesis would not have been possible I would like to express my sincere gratitude to Dr Brian Hopkins for his unwavering support and guidance throughout the study and never losing hope Dr Hopkins insight and patience have truly been essential to the completion of this project and I thank him immensely I would also like to thank the Mathematics Department at Saint Peter s University for giving me the tools to pursue this project I am also grateful for the love and reassurance my family gives me everyday Without their encouragement, I would have never been able to pursue my interests at Saint Peter s University Even though they reside half a world away, their counsel has remained an important part of my life Last, but certainly not least, I would like to thank Dr Rachel Wifall and the Honors Program at Saint Peter s University for giving me the opportunity to venture out on this research experience, which has certainly put my time here at Saint Peter s to the test 2

4 Contents 1 Introduction 4 11 Finite Dynamical Systems 4 12 Partitions 4 13 Bulgarian Solitaire Row Bulgarian Solitaire Operation 9 2 Block Notations Row Bulgarian Solitaire on Single Blocks Cycles in Block Notation 17 3 Characterizing Cyclic Partitions Cycles with Zero Incomplete Diagonals Cycles with One Incomplete Diagonal Cycles with Two Incomplete Diagonals Cycles with Three Incomplete Diagonals Impossibility of Four or More Incomplete Diagonals 27 4 Enumeration of Cyclic Partitions Counting with Zero or One Incomplete Diagonals Counting with Two Incomplete Diagonals Diagonals Diagonals Diagonals Counting with Three Incomplete Diagonals Examples and data 33 3

5 1 Introduction 11 Finite Dynamical Systems Given a nonempty finite set X, a finite dynamical system is the pair px, φq such that φ : X Ñ X A repeated composition of the operation φ to itself is referred to as an iterations Given α P X and n P N, an n-fold composition of φ is denoted by φ n pαq pφ φ φqpαq The set X partitions into two subsets when φ is iterated across all elements in X: the cyclic set and the tail set The cyclic set, CycpX, φq of X are a part of a cycle, ie, CycpX, φq tα P X : φ n pαq α for some n P Nu The tail set, denoted by TailpX, φq, consists of elements that do not form cycles when iterated by φ Given x P X, if m is the least non-negative integer such that φ m pxq is cyclic, then φ k pxq P TailpX, φq for any 0 ď k ă m The ends of a tail, that is, any z P X such that there is no y P X with φpyq z, are called a Garden of Eden states (from the language of cellular automata) 12 Partitions Simply stated, a partition of an integer n is a way writing n as the sum of positive integers summands Since order does not matter, partitions are conventionally written in nonincreasing order of summands For example, there are 7 partitions of five given below: 5 4 ` 1 3 ` 2 3 ` 1 ` 1 2 ` 2 ` 1 2 ` 1 ` 1 ` 1 1 ` 1 ` 1 ` 1 ` 1 A partition can be denoted (2,1,1,1), or using short-hand notation, 2111, or even more succinctly as 21 3 In general, a partition λ pλ 1, λ 2,, λ t q with integers λ 1 ě λ 2 ě ě λ t ą 0 is a partition of n ř t i 1 λ i with t parts We denote the set of all partitions of n by P pnq 4

6 We use the Ferrers diagram to represent partitions as an array of dots For example, the partition has Ferrers diagram Each summand λ i corresponds to a column of λ i dots In this paper, we will see that Ferrers diagram are helpful tool for understanding some operations on partitions 13 Bulgarian Solitaire Bulgarian solitaire combines finite dynamical system with integer partitions It was first introduced by Brandt in 1982 [3] Given a partition λ pλ 1,, λ t q, Bulgarian solitaire is defined by the operation α : P pnq Ñ P pnq where αpλq tt, λ 1 1, λ 2 1,, λ t 1u The parts of the image may need to be reordered to be nonincreasing In terms of the Ferrers diagram, it takes the bottom row of dots and turns it into a column while the rest of the partitions remain unchanged [4] To understand the operation better, Here are some examples of Bulgarian solitaire operation on partitions of 5 The black dots represent the changes in the Ferrers diagram with each Bulgarian solitaire application 1 αpp1, 1, 1, 1, 1qq p5q

7 2 αpp5qq p4, 1q αpp4, 1qq p3, 2q Note: There is a reordering of parts where t is placed after λ 1 1, since t 2 ă 3 λ αpp3, 2qq p2, 2, 1q αpp2, 2, 1qq p3, 1, 1q αpp3, 1, 1qq p3, 2q

8 7 αpp2, 1, 1, 1qq p4, 1q The full map of the Bulgarian solitaire operation on P p5q is Ó Ó Ö Ò 5 Ñ 41 Ñ 32 Ñ 221 The figure above is an example of a finite dynamic system under the Bulgarian solitaire operation TailpP p5q, αq tp5q, p4, 1q, p2, 1, 1, 1q, p1, 1, 1, 1, 1qu Of these, p1, 1, 1, 1, 1q and p2, 1, 1, 1q are Garden of Eden states since they have no pre-images CycpP p5q, αq tp3, 2q, p2, 2, 1q, p3, 1, 1qu We observe that all partition of P pnq eventually form cyclic partitions The cyclic partitions for Bulgarian solitaire have been intensively studied in [3, 4, 5, 6] Bulgarian solitaire is a variation of one of the basic operations on integer partition, conjugation Given the partition λ pλ 1,, λ t q, the conjugate partition λ 1 is defined as λ 1 pλ 1 1,, λ 1 sq where λ 1 is the number of parts λ i greater than or equal to i [1] Take for example the conjugation of partition p2, 2, 1q According to the definition, when i 1, there are three parts of p2, 2, 1q that are greater or equal to i, therefore λ 1 3 Note that λ 1 1 is equal to t, the number of parts (since every summand is greater than or equal to 1) When i 2, only 2 parts of p2, 2, 1q are greater or equal to i, thus λ 2 2 There are no parts greater or equal to i 3, therefore, we stop Thus, p2, 2, 1q 1 p3, 2q Similarly, p3, 2q 1 p2, 2, 1q The conjugate of a partition is a reflection of the partition s Ferrers diagram along the block diagonal (or y x line), swapping the rows and 7

9 columns Looking at the Ferrers diagram of the conjugation of p2, 2, 1q and p3, 2q, we observe that all the rows of the first partition p2, 2, 1q are simple converted into columns in the second partition, p3, 2q Bulgarian solitaire can be viewed as a first step towards conjugation, in that one row is converted to a column The next step, converting two rows to columns, is the operation studied in this thesis, defined more formally in the next section See [4] for some results on this family of operations A cycle partition λ for Bulgarian solitaire has the form pk ` δ k, k 1 ` δ k 1,, 2 ` δ 2, 1 ` δ 1, δ 0 q where each δ i is either 0 or 1 [3] For cases where δ i is 0 for all 1 ď i ď k, we see that the partitions is in the form pk, k 1,, 2, 1q whose sum is kpk `1q{2 These types of numbers are called triangular number denoted by T k All triangular number T k form a cyclic partition λ in the form pk, k 1,, 3, 2, 1q under Bulgarian solitaire [2] Any partition λ in the form pk, k 1,, 2, 1q does not change under Bulgarian solitaire The partition p3, 2, 1q of 6 is shown below α ý This builds our study to explain characteristics for partitions where δ i is not 0 for all 1 ď i ď k Assuming a nonnegative integer n, write n T k ` r where k is chosen to be the as large as possible for a nonnegative r Here, r is equal to δ k ` δ k 1 ` ` δ 2 ` δ 1 ` δ 0 The cyclic partition under Bulgarian solitaire for n will have an unchanging triangular array of T k dots in the form pn, n 1,, 2, 1q The variation comes from the remaining r dots which are placed in the pk ` 1qst diagonal [3] The total number of ways to place r 8

10 dots in k ` 1 spaces is simply given by `k`1 r A 2-element cycle between partitions of 18 is shown below α And a 4-cycle among the partitions of 7 α α α Row Bulgarian Solitaire Operation The 2-row Bulgarian solitaire or β operation for a partition λ pλ 1,, λ t q is defined as βpλq tλ 1 1, λ 1 2, λ 1 2,, λ t 2u, where λ 1 i is the number of parts of λ ě i As before, the parts of the image are rearranged in nonincreasing order starting from the left to right In terms of the Ferrers diagram, the β operation takes the bottom two rows of the partition and changes them to columns, then rearranges the parts so that the order is non increasing Here are several examples of 2-row Bulgarian solitaire 9

11 1 βpp1, 1, 1, 1, 1qq p5q Note: Because there is only one row, the effects of β is the same as α and conjugation 5 2 βpp5qq p3, 1, 1q Note: There is a reordering of parts in this step: since λ 1 1 λ 1 2 ă λ 1 2, ie, 1 ă 3, the image is pλ 1 2, λ 1 1, λ 1 2q or p3, 1, 1q 3 βpp3, 1, 1qq p3, 1, 1q Note: The image produced by β is the same as conjugation 4 βpp2, 1, 1, 1qq p4, 1q

12 Note: The image produced by β is the same as α and conjugation 5 βpp4, 1qq p2, 2, 1q Note: The parts have been rearranged to pλ 1 1, λ 1 2, λ 1 2q 6 βpp2, 2, 1qq p3, 2q Note: Image produced by β is the same as conjugation 7 βpp3, 2qq p2, 2, 1q Note: The image produced by β is the same as conjugation Like Bulgarian solitaire, the β operation also forms a finite dynamical system on integer partitions The full map of the 2-row Bulgarian Solitaire operation on P p5q is given below: Ó Ó Ö 5 Ñ 311 ý 41 Ñ 221 The β map for P p5q has TailpP p5q, βq tp1, 1, 1, 1, 1q, p5q, p2, 1, 1, 1q, p4, 1qu and CycpP p5q, βq tp3, 1, 1q, p3, 2q, p2, 2, 1qu Out of the Tail elements, 11

13 p1, 1, 1, 1, 1q and p2, 1, 1, 1q are Garden of Eden states The Garden of Eden states for β (and the entire family of operations between Bulgarian solitaire) are characterized in [4] The focus of this paper is characterizing cyclic partitions of β for all n 12

14 2 Block Notations To study cyclic partitions under the β operation more effectively, we use a slightly different notation to represent partitions We divide the Ferrers diagram into 2 ˆ 2 blocks and represent them via the sum of the dots within the each 2 ˆ 2 block A complication arises when summing the blocks with 2 dots Since orientation matters, label the vertically aligned two dots as V and the horizontally aligned 2 dots as H; when mentioned together, we call these 2-blocks Block notation allows only certain combination of block to ensure that the underlying summands are nonincreasing and there are are no gaps within any particular column of the Ferrers diagram which is discussed later in this section In other words, only certain combinations of blocks correspond to partitions All possible types of blocks are shown below 1-block H-block V -block 3-block 4-block A 1-block has three empty spaces within its block We cannot have any block to the right of a 1-block to preserve the nonincreasing rule of partitions In addition, due to the gap in the top half of the 1-block, it cannot support any block above it A H-block has two empty spaces in the top half of its block Thus, to prevent any gaps, the H-block is unable to support block above it However, the dot in the lower right makes the H-block more flexible than the 1-block: an H-block can support either a 1-block or another H-block to the right These configurations of H-blocks are given below H 1 H H 13

15 A V -block also has two empty spaces like the H-block However, unlike the H-block, these empty spaces are vertically aligned to the right half of the block Thus, while the V -block is able to support a 1-block or another V -block above it, it cannot support anything to the right These configurations of V -blocks are given below 1 V V V A 3-block only has one empty space to the top right of the block It is a superposition of the H- and V - block; hence, is able to support any aforementioned configuration of 1 and V -block above itself or 1 and H-block to the right The 3-block cannot support another 3- or 4-block above or to the right due to an empty space on the top right corner Some configurations with a 3-block are given below: H V 3 A 4-block is the only block that is completely filled This makes the 4- block gap free, thus provides a gap-less base upon which can support any of the aforementioned configuration of blocks including another 4-block This also solves the problem of placing blocks in a non increasing order Since the 4-block has its right column filled, it can also support any block to the right 14

16 This characteristic of 4-blocks plays an important role when talking about cyclic partitions Let us consider the partition (6,3,3,3,1) Now, divide the number of dots by 2 ˆ 2 block starting from the bottom left of the Ferrers diagram as shown in the figure below left The equivalent block diagram is given on the right V 3 H The block notation, motivated by β affecting two rows, will be superior to Ferrers diagrams as we analyze cyclic partitions We can also define block notation without reference to the Ferrers diagrams Given a partition λ pλ 1,, λ t q, the summands λ 2i 1 and λ 2i determine the blocks for the ith column If the number of parts t is odd, we add a part λ t`1 0 so that parts can be considered in pairs If λ 2i is even, then the ith column of the block diagram will have λ 2i {2 4-blocks If the difference between λ 2i 1 and λ 2i is even then tpλ 2i 1 λ 2i q{2u 15

17 V -blocks are to added on top of any 4-blocks in the ith column If the difference is odd, we add tpλ 2i 1 λ 2i q{2u V -blocks and a 1-block above the 4-blocks in the ith column For example, partition (5,2) has λ 1 5, λ 2 2, and i 1 We have λ 2 {2 2{2 1 4-block We have tpλ 1 λ 2 q{2u tp5 2q{2u t3{2u 1 V -block Since pλ 1 λ 2 q 3 1 mod 2, we add a 1-block to the top of the first column Thus, partition (5,2) has block notation 1 V 4 If λ 2i is odd, then the ith column will have pλ 2i 1q{2 4-blocks If λ 2i 1 λ 2i, then add one H-block to the ith column If λ 2i 1 ą λ 2i, then add one 3-block and tpλ 2i 1 λ 2i 1q{2u V -blocks Further, if λ 2i 1 λ 2i 1 is odd, then add a 1-block at the top of the ith column For example, partition p7, 3q has λ 1 7, λ 2 3 and i 1 We have pλ 2 q{2 p3 1q{2 1 4-block Since λ 1 ą λ 2, ie, 7 ą 3, we have one 3-block We have tpλ 1 λ 2 1q{2u tp7 3 1q{2u t3{2u 1 V -block Since λ 1 λ mod 2, we also have one 1-block Thus, partition (7,3) has block notation 16

18 1 V 3 4 Block notation also gives rise to a concept of column distance which will be very important when dealing with cyclic partitions with multiple incomplete diagonals Simply stated, the column distance k from a reference block r to a target block t is the number of columns separating r from t In the block notation of partition p6, 3, 3, 3, 1q above, the 3-block in the first column is two columns away from the 1-block in the third column, thus it has a column distance of Row Bulgarian Solitaire on Single Blocks Before we analyze cyclic partition, it is important to understand how the β operations affects a single block When acting on a single block, the β operation is equivalent to conjugation In other words, β acting on a single block is the same as reflecting the Ferrers diagram across the line y x Due to their symmetry across the block diagonal, the 1-, 3-, and 4-blocks remain unchanged under the β operation The H-blocks and V -blocks, however, are not symmetric along the block diagonal, so they are not fixed by β Instead, βpβp1, 1qq βpp2qq p1, 1q That is, the H-blocks and V -blocks are swapped by β This is in fact the source of most of the complexity surrounding the characterization problem of this study H V 22 Cycles in Block Notation Using block notations, characterizing the cyclic partitions of β operations becomes much easier It helps us to see the intricate changes that are not as 17

19 apparent in the Ferrers diagrams Let us take a look at the β cycle of (5,4,2,2) Using the block notation, we can see a much clearer pattern in cyclic partitions: The block notation shows us that the 1-block simply descends along a diagonal over the 4-blocks, as with the cyclic partitions of Bulgarian solitaire We shall expand this concept of block diagonals in the next section In contrast to Bulgarian solitaire, where we considered diagonals of dots, in 2-row Bulgarian solitaire, we need to look at the problem using diagonals of 2 ˆ 2 dots The ith block diagonal is composed of i blocks that play an important role to distinguish cyclic partitions from tail partitions The diagonals in 2-row Bulgarian solitaire can be divided into two categories: complete diagonals and incomplete diagonals 18

20 A complete diagonal consists entirely of 4-blocks, leaving no empty spaces within the diagonal When dealing with cyclic partitions, we will see that complete diagonals form a triangular array of 4-blocks that remains fixed by β, much like the triangular array of dots in Bulgarian solitaire The variation in cyclic partitions for 2-row Bulgarian solitaire β arises from subsequent diagonals, ie, incomplete diagonals In contrast to complete block diagonals, any diagonal that is not completely filled with 4-blocks forms an incomplete diagonal The name is attributed to the empty spaces in the blocks that make the block diagonal In the previous section, we looked at the β cycle of p5, 4, 2, 2q We observe that the first and second diagonals are complete, since they are completely filled with 4-blocks while the outermost third diagonal is incomplete, consisting of one 1-block Let us look at another example of partition p5, 4, 4, 1, 1, 1q to better understand complete and incomplete diagonals The d values to the left of the Ferrers diagram and block notation denote the dth diagonal d 3 d 2 d 1 d 3 1 d 2 4 V d H We see that the first diagonal, d 1, is the only diagonal complete, since the one 4-block that fills the first diagonal The second diagonal, d 2, has only one 4-block among its two blocks, thus is incomplete Similarly, since the third diagonal, d 3, is also not completely filled with three 4-blocks, it is also incomplete 19

21 Cyclic partitions operating under β have the property that blocks stay in the same diagonal Specifically, the d blocks of the dth diagonal cycle, except that any V - and H-blocks swap when moving from the bottom row to the first column In any cyclic partitions, we will show that there may be up to three diagonal cycles operating under β The length of any particular β cycle can be determined by taking the lowest common multiple of the total amount of steps taken for each diagonal to cycle with a possible factor of 2 when 2-blocks are present 20

22 3 Characterizing Cyclic Partitions Partitions with only complete block diagonals provide very few cyclic partitions for β Many more arise with one incomplete diagonal, similar to α There are some cyclic partitions, though, that have two or three incomplete diagonals We treat the number of incomplete diagonals in turn and prove that there cannot be four or more in a cyclic partition 31 Cycles with Zero Incomplete Diagonals Lemma 1 If a partition λ of n has zero incomplete diagonals then n 4T k for some k and λ p2k, 2k,, 4, 4, 2, 2q Proof Having only complete diagonals, say k of them, means the block diagram of λ consists of k diagonals full of 4-blocks Thus n 4T k and λ must have the desired form Lemma 2 If a partition λ of n has zero incomplete diagonals, then λ is cyclic and has cycle length one Proof From Lemma 1, we know λ p2k, 2k,, 4, 4, 2, 2q, with 2k parts Thus, the β image of λ is βpp2k, 2k,, 4, 4, 2, 2qq p2k, 2k, 2k 2, 2k 2,, 4 2, 4 2, 2 2, 2 2q p2k, 2k, 2k 2, 2k 2,, 2, 2q λ In other words, β maps λ to itself Hence, it has cycle length one 32 Cycles with One Incomplete Diagonal Cyclic partitions with one incomplete diagonal have no restrictions as to the content of that diagonal, since every block is supported horizontally and vertically by a 4-block 33 Cycles with Two Incomplete Diagonals We will see that there are three different types of cyclic partitions with two incomplete diagonals, which we consider in turn In general, an m-n diagonal means that the lower incomplete diagonal has an m-block as its smallest block the the higher diagonal has an n-block its largest block 21

23 By a 2-1 diagonal we mean that there is at least one type of 2-block on the ith diagonal along with larger blocks and at least one 1-block on the pi ` 1qst diagonal with no larger blocks Lemma 3 Suppose a cyclic partition λ with incomplete diagonals i and i`1 satisfies the 2-1 diagonal description Then i must be odd and there are two conclusions about the content of the incomplete diagonals: The 1-blocks in the pi ` 1qst diagonal must appear in alternating positions at even distance from one another In the ith diagonal, any V -blocks are at an even distance from the 1- blocks while all H-blocks are at an odd distance from the 1-blocks Proof Consider a partition λ with a triangular array of T i 1 4-blocks, an ith diagonal with one V -block in the first column and i 1 4-blocks filling the rest of the ith diagonal, and an pi ` 1qst diagonal with a single 1-block directly on top of the V-block We an represent this as d i ` 1 d i d i 1 d 1 1 V After i 2 iterations of β, the partition will have arrangement β 0 d i ` 1 1 d i d i β Ñ H d V β i2 β i2`1 22

24 After this arrangement, at the pi 2 ` i ` 1qst iteration, the cycle goes back to the original partition d i ` 1 d i d i 1 d 1 1 V β i2`i`1 We observe in this cycle that all the H-blocks occur an odd distances from the 1-block while the V - blocks occur only at even distances from the 1-block The restrictions are the same when there are aditional 2-blocks on the ith diagonal The pi ` 1qst diagonal can have 1-blocks in every other position at most A 3-2 diagonal refers to a cyclic partition with at least one 3-block on the ith diagonal along with 4-blocks and at least one 2-block on the pi ` 1qst diagonal along with 1-blocks Lemma 4 Suppose a cyclic partition λ with incomplete diagonal i and i ` 1 satisfies the 3-2 diagonal description Then i must be even and there are two conclusions about the content of the incomplete diagonals: The H- and V -block in the pi ` 1qst diagonal must appear at an even distance from like 2-blocks and an odd distance from unlike 2-blocks In the ith diagonal, any 3-block is at an even distance from any V -blocks and an odd distance from any H-blocks Proof Consider a partition λ with a triangular array of T i 1 4-block, an ith diagonal with one 3-block in the first column and 4-blocks filling the rest of the ith diagonal, and an pi ` 1qst diagonal with a V -block directly on top of the 3-block Thus, λ is 23

25 d i ` 1 d i d i 1 d 1 V β 0 After i 2 iterations of β, the partition will be d i ` 1 H d i d i β Ñ 3 H 4 4 d β i2 β i2`1 After this, at pi 2 ` i ` 1qst iteration, the cycle returns to the original partition d i ` 1 d i d i 1 d 1 V β i2`i`1 We observe in this cycle that all the H-blocks occur an odd distances from the 3-block while the V -blocks occur only at even distances from the 3-block The restrictions are the same when there are additional 3-blocks on the ith diagonal in at most every other position and additional 2-blocks 24

26 in the pi ` 1qst diagonal There are no restrictions for any 1-blocks in the pi ` 1qst diagonal The last type of two incomplete diagonals is the 3-1, refers to a cyclic partitions with at least one 3-block on the ith diagonal along with 4-blocks and at least one 1-block on the pi ` 1qst diagonal with no larger parts Since a 1-block is supported by a 3-block, and these blocks themselves do not change under β, due to their symmetry, there are no diagonal nor positional restrictions for this type of two incomplete diagonal 34 Cycles with Three Incomplete Diagonals The three incomplete diagonal case can be reduced to the study of two overlapping two incomplete diagonal cases, namely the 3-2 and the 2-1 diagonal cases Lemma 5 Suppose a cyclic partition λ with incomplete diagonals i, i ` 1, and i ` 2 has at least one 3-block on the ith diagonal along with 4-blocks, 2-blocks on the pi ` 1qst diagonal, and at least one 1-block on the pi ` 2qnd diagonal with no larger blocks Then i must be even and there are three conclusions about the content of the incomplete diagonals: The pi ` 1qst diagonal is completely filled with alternating H- and V - blocks The 1-blocks in the pi ` 2qnd diagonal must appear in alternating positions at even distances from one another In addition, all 1-blocks must be an even distance from the V -blocks and an odd distance from the H-blocks In the ith diagonal, any 3-block must appear in alternating positions at even distances from one another In addition, all 3-blocks must be an even distance from the V -blocks and an odd distance from the H-blocks Proof Let us take a partition λ with a triangular array of T i 1 4-blocks, an ith diagonal with one 3-block in the first column and i 1 4 -blocks, an pi ` 1qst diagonal with alternating H- and V - blocks, and an pi ` 2qnd diagonal with one 1-block in the first column Thus λ is 25

27 d i ` 2 d i ` 1 d i d i 1 d 1 1 V 3 H 4 4 V V After i 2 {2 1 iterations of β, the partition will be β 0 d i ` 2 1 d i ` 2 d i ` 1 V d i ` 1 H d i d i 1 4 H 4 β pi`1q d i d i 1 4 V 4 3 H 4 V d V d H 1 β pi2 {2q 1 β pi2 {2q`i After this, the pi 2 {2`i`1qst iteration takes the cycle back to the original partition d i ` 2 d i ` 1 d i d i 1 d 1 1 V 3 H 4 4 V V β pi2 {2q`i`1 26

28 We observe in this cycle that all the H-blocks occur an odd distances from the 3-block while the V - blocks occur only at even distances from the 3-block The restrictions are the same when there are additional 3-blocks in at most every other position of the ith diagonal and additional 1-blocks in at most every other position of the pi ` 2qnd diagonal 35 Impossibility of Four or More Incomplete Diagonals Lemma 6 Cyclic partition are limited to no more than three incomplete diagonals Proof From the previous lemma, we know that three incomplete diagonals is possible only for an pi`2qnd even diagonal of alternating 1-blocks completely supported by an pi ` 1qst diagonal of alternating H and V blocks Similarly, the pi ` 1qst diagonal with alternating H- and V -blocks is supported by an incomplete ith diagonal with at least one 3-block However, we know that a 3-block can only be supported by a 4 block, so the pi 1qst diagonal has to be complete for there to be a cycle Also, the pi ` 2qnd diagonal with at least one 1-block cannot support anything on the next diagonal Thus, there can be no more than three incomplete diagonals 27

29 4 Enumeration of Cyclic Partitions We complete the thesis with results on counting cyclic partitions under β 41 Counting with Zero or One Incomplete Diagonals For cyclic partitions of n, write n 4T i 1 ` r where i is chosen to be as large as possible for r nonnegative If r 0, there are no incomplete diagonals, and there is exactly one cyclic partition, described above Therefore, assume 1 ď r ď 4i 1 We will derive a formula for the number of ways these r dots can be arranged starting in the ith diagonal beginning with just 1-blocks and adding larger blocks until all possibilities are incorporated Suppose the r dots are distributed in r 1-blocks on the ith diagonal (this is only sensible for such that r ď i) This can be done in `i r ways (using a binomial coefficient or choose number ) Now, let us assume the r dots can be distributed by 2- and 1-blocks The number of these combinations is t r 2 u ÿ j 0 2 jˆi ˆ i j j r 2j In this formula, j is the number of 2-blocks The 2 j factor accounts for each 2-block being either a V -block or an H-block Note that the position of H- and V -blocks does not matter here since there is only one incomplete diagonal The first binomial coefficient counts the ways to place the j 2-blocks among the i positions of the ith diagonal This accounts for 2j of the r dots The second binomial coefficient chooses which of the remaining i j positions on the diagonal are filled with the r 2j 1-blocks necessary to place all r dots The reasoning is similar when we add the possibility of k 3-blocks: t r 3 u ÿ k 0 ˆi k t r 3k 2 u ÿ j 0 2 jˆi ˆ k i j k j n 2j 3k 28

30 Finally, we allow l 4-block in the ith block diagonal We rewrite the resulting formula more concisely using a multinomial coefficient: t r 4 u t r 4l u t pr 4l 3iq u 3 2 ÿ l 0 ÿ k 0 ÿ j 0 ˆ 2 j i l, k, j, r 2j 3k 4l, i r ` 3l ` 2k ` j Working from the outside in, the limits of summation come from the maximum possible number of 4-, 3-, and 2-blocks, respectively The multinomial coefficient should be understood as follows: the i positions of the ith diagonal are to be filled with l 4-blocks, k 3-blocks, j various 2-blocks, and r 2j 3k 4l 1-blocks, leaving i l k j pr 2j 3k 4lq i r`3l`2k`j empty positions 42 Counting with Two Incomplete Diagonals For cyclic partition of n with two incomplete diagonals, as mentioned earlier there are three variant Write n 4T i 1 ` r where i is chosen to be as large as possible for a nonnegative r The types of diagonals depends on whether i is even or odd Here, r is the number of dots in the outermost diagonal in a one incomplete diagonal arrangement For an even i, If 7 i ` 2 ď r ď 6i ` 1, then n has a 3-2 diagonal case 2 If 3i ` 1 ď r ď 5i, then n has a 3-1 diagonal case For an odd i, If 2i ` 1 ď r ď 9i 3, then n has a 2-1 diagonal case 2 If 3i ` 1 ď r ď 5i, then n has a 3-1 diagonal case If r does not meet any of these criteria, this means that there are not enough dots to make two incomplete diagonals Thus we use dots from the outermost complete diagonal to make the two incomplete diagonals To do 29

31 this, we Write n 4T i ` m where i is chosen to be as large as possible for a nonnegative m The only difference here is we set r n 4T i 1 After that like we did previously, depending on whether i is even or odd, we use the previous criteria to see the diagonals n can accommodate depending the value of the new r Diagonals Let us first look at the case 3-1 diagonal case Assume there are k 3-blocks arranged in i positions, while the remaining i k positions on the diagonal are filled by 4-blocks The number of ways to arrange this is given by `i k, which accounts for the 3k ` 4pi kq 4i k of the r dots The remaining r p4i kq r ` k 4i dots are arranged as 1-blocks in the pi ` 1qst diagonal, leaving i ` 1 pr ` k 4iq 5i r k ` 1 positions empty, given by ` i`1 r`k 4i Thus, we may count all 3-1 diagonal cyclic partitions using the formula iÿ k 1 ˆ ˆi i ` 1 δpr ` k 4i ą 0q k r ` k 4i The Kronecker delta factor δppq is 1 if p is true, 0 if p is false This is also known as the Iverson bracket Thus, if there are fewer dots than these blocks can accommodate with at least one 3-block in the ith diagonal and i ` 1 1-blocks in the pi ` 1qst diagonal, then the delta function reduces the formula to zero since such arrangements are not valid Diagonals For the 2-1 diagonal case, as mentioned earlier, i must be odd As Lemma 3 states, assume the ith diagonal has j 2-blocks with 1 ď j ď i The remaining i j position are filled by k 3-blocks and i j k 4-blocks The number of ways this can be done is given by the product `i `i j j k and 30

32 accounts for 2j ` 3k ` 4pi j kq 4i 2j k dots The remaining r 2j 3k 4pi j kq r ` 2j ` k 4i dots are arranged as 1-blocks in pi ` 1q{2 positions in the pi ` 1qst diagonal, leaving pi ` 1q pr ` 2j ` k 4iq 5i 2j k r ` 1 empty spaces The number of these combinations is ` pi`1q{2 r 4i`2j`k Thus, the formula iÿ i j ÿ j 1 k 0 ˆi ˆi j 2 j k ˆ pi ` 1q{2 δpr 4i ` 2j ` k ą 0q r 4i ` 2j ` k gives the total number of ways we can arranges r dots in a 2-1 diagonal case In the formula, the factor two accounts for the two sets of way the H- and V -blocks can be arranged for each configuration If there are fewer dots than these blocks can accommodate with at least one 2-block in the ith diagonal and i ` 1 1-blocks in the pi ` 1qst diagonal, then the delta function reduces the formula to zero since such arrangements are not valid Diagonals For a 3-2 diagonal, i must be even as mention in Lemma 4 As Lemma 4 states, assume the ith diagonal has k 3-blocks with 1 ď k ď i{2 The remaining i k position are filled by 4-blocks The number of ways this can be done is given by `i k This accounts for 3k `4pi kq 4i k dots The remaining r p4i kq r 4i ` k dots are in the pi ` 1qst diagonal as least one 2-block, counted by j, and r 4i ` k 2j 1-blocks There are i ` 1 j pr 4i ` k 2jq 5i ` j r k ` 1 empty positions in the pi ` 1qst diagonal The number of ways these arrangements are made can be written as ` i`1 j r 4i`k 2j Thus, the formula 31

33 i 2ÿ i`1 ÿ k 1 j 1 ˆi 2 k ˆ ˆi ` 1 j i ` 1 j r 4i ` k 2j gives the total number of ways we can arrange r dots in the 3-2 diagonal case In the formula, the 2 accounts for the two sets of ways H- and V -blocks can be arranged for each configuration 43 Counting with Three Incomplete Diagonals For cyclic partitions of n, write n 4T i ` m where i is chosen to be as large as possible for a nonnegative m For a diagonal case, i must be even The pi 1qst diagonal will be the last complete diagonal Thus, the number dots that make up the three diagonals r is given by r n 4T i 1, where 3 ` 11 2 i ď r ď 2 ` 13 2 i According to Lemma 5, the ith diagonal must have at least one 3-block, the pi`1qst diagonal must have alternating H- and V -blocks, and the pi`2qnd diagonal must have at least one 1-block Let the ith diagonal have k 3-blocks, while the remaining positions are filled with i k 4-blocks The number of ways these blocks can be arranged is `i{2 k Similarly, the alternating H- and V -blocks in the pi ` 1qst diagonal have only 2 possible combinations The ith and pi ` 1qst diagonal account for 3k ` 4pi kq ` 2pi ` 1q 6i k ` 2 dots The remaining r p6i k ` 2q r 6i ` k 2 dots are arranged as 1-blocks in alternating positions of the pi`2qnd diagonal with pi`2q pr 6i`k 2q 7i r k `4 empty spaces; possible configurations are counted by ` pi`2q{2 r 6i`k`2 Thus the formula i 2ÿ k 1 ˆi{2 k ˆ pi ` 2q{2 2 δpr 6i ` k ` 2 ą 0q r 6i ` k ` 2 gives the total number of way we can arrange r dots in a diagonal case If there are fewer dots than these blocks can accommodate with at least one 3-block in the ith diagonal, alternative H-and V -block for the pi ` 1qst 32

34 diagonal and at least one 1-blocks in the pi ` 2qnd diagonal, then the delta function reduces the formula to zero since such arrangements are not valid 44 Examples and data Here we give some examples of cyclic partitions under β and present a table of numbers of cyclic partitions by type up to n 25, the values we were able to able to verify through other means using Mathematica Here is an example of a length 4 cycle among the partitions of Ò ÐÝ Ó This is clearer when the same partitions are shown in block notation 33

35 1 V 3 H 4 4 V H 1 4 V 4 3 H Ò H 4 V 4 3 H 1 ÐÝ Ó V 3 H V These are the smallest partitions with three incomplete diagonals arises in partitions of 18, shown first in standard notation This leaves r dots to place Applying the formula derived for three incomplete diagonals in this case corroborates the number we have found: ˆ1 1 ˆ ` 1 ` 4 2 ˆ The following table shows the number of cyclic partitions for 1 ď n ď 25 by number of incomplete diagonals and, in the case of two incomplete diagonals, the three types we have characterized izn : : : Σ

36 izn : : : Σ

37 References [1] George Andrews, The Theory of Partitions, Cambridge University Press, 1998 [2] Hans-Joachim Bentz, Proof of the Bulgarian solitaire conjectures, Ars Combin 23 (1987) [3] Jørgen Brandt, Cycles of partitions, Proc Amer Math Soc 85 (1982) [4] Brian Hopkins, Column-to-row operations on partitions: The envelopes, Integers 9 Supplement (2009), A6 [5], 30 years of Bulgarian solitaire, College Math J 43 (2012) [6] Brian Hopkins and Michael A Jones, Shift-induced dynamical systems on partitions and compositions, Electron J Combin 13 (2006) R80 36